Solveeit Logo

Question

Question: An ellipse intersects the hyperbola \(2{x^2} - 2{y^2} = 1\) orthogonally. The eccentricity of the el...

An ellipse intersects the hyperbola 2x22y2=12{x^2} - 2{y^2} = 1 orthogonally. The eccentricity of the ellipse is reciprocal of that of the hyperbola. If the axis of the ellipse are along the coordinate axes, then
(a)\left( a \right) Equation of ellipse is x2+2y2=2{x^2} + 2{y^2} = 2
(b)\left( b \right) The foci of ellipse are (±1,0)\left( { \pm 1,0} \right)
(c)\left( c \right) Equation of ellipse is x2+2y2=4{x^2} + 2{y^2} = 4
(d)\left( d \right) The foci of ellipse are (±2,0)\left( { \pm \sqrt 2 ,0} \right)

Explanation

Solution

In this particular question use the concept that if the equations a1x2+b1y2=1,a2x2+b2y2=1{a_1}{x^2} + {b_1}{y^2} = 1,{a_2}{x^2} + {b_2}{y^2} = 1 intersects orthogonally then the condition is 1a11b1=1a21b2\dfrac{1}{{{a_1}}} - \dfrac{1}{{{b_1}}} = \dfrac{1}{{{a_2}}} - \dfrac{1}{{{b_2}}} and use the concept that the relation between, eccentricity e, semi major axis a, and semi minor axis b in case of hyperbola is given as b2=a2(e21){b^2} = {a^2}\left( {{e^2} - 1} \right) so use these concepts to reach the solution of the question.

Complete step-by-step solution:
Given data:
Given equation of hyperbola
2x22y2=12{x^2} - 2{y^2} = 1
Let the equation of ellipse is x2a2+y2b2=1\dfrac{{{x^2}}}{{{a^2}}} + \dfrac{{{y^2}}}{{{b^2}}} = 1
Now as we know that that if the equations a1x2+b1y2=1,a2x2+b2y2=1{a_1}{x^2} + {b_1}{y^2} = 1,{a_2}{x^2} + {b_2}{y^2} = 1 intersects orthogonally then the condition is 1a11b1=1a21b2\dfrac{1}{{{a_1}}} - \dfrac{1}{{{b_1}}} = \dfrac{1}{{{a_2}}} - \dfrac{1}{{{b_2}}}................ (1)
So on comparing with equation of ellipse,
a1=1a2,b1=1b2\Rightarrow {a_1} = \dfrac{1}{{{a^2}}},{b_1} = \dfrac{1}{{{b^2}}}
And on comparing with equation of hyperbola we have,
a2=2,b2=2\Rightarrow {a_2} = 2,{b_2} = - 2
Now from equation (1) we have,
a2b2=12(12)=12+12\Rightarrow {a^2} - {b^2} = \dfrac{1}{2} - \left( { - \dfrac{1}{2}} \right) = \dfrac{1}{2} + \dfrac{1}{2}
a2b2=1\Rightarrow {a^2} - {b^2} = 1................... (2)
Now as we know that the relation between, eccentricity e3{e_3}, semi major axis a3{a_3}, and semi minor axis b3{b_3} in case of hyperbola is given as b32=a32(e321)b_3^2 = a_3^2\left( {e_3^2 - 1} \right).
Now the standard equation of hyperbola is x2a32y2b32=1\dfrac{{{x^2}}}{{a_3^2}} - \dfrac{{{y^2}}}{{b_3^2}} = 1 now compare it with given equation of hyperbola 2x22y2=12{x^2} - 2{y^2} = 1 we have,
a32=12,b32=12\Rightarrow a_3^2 = \dfrac{1}{2},b_3^2 = \dfrac{1}{2}
Therefore, b32=a32(e321)b_3^2 = a_3^2\left( {e_3^2 - 1} \right) become,
12=12(e321)\Rightarrow \dfrac{1}{2} = \dfrac{1}{2}\left( {e_3^2 - 1} \right)
1=(e321)\Rightarrow 1 = \left( {e_3^2 - 1} \right)
e32=2\Rightarrow e_3^2 = 2
e3=2\Rightarrow {e_3} = \sqrt 2
Now it is given that the eccentricity of the ellipse is reciprocal of that of the hyperbola.
So e1=1e3=12{e_1} = \dfrac{1}{{{e_3}}} = \dfrac{1}{{\sqrt 2 }}, where e1{e_1} is the eccentricity of the ellipse.
Now as we know that the relation between, eccentricity e, semi major axis a, and semi minor axis in case of ellipse is given as b2=a2(1e12){b^2} = {a^2}\left( {1 - e_1^2} \right).
b2=a2(1(12)2)=a2(112)=a22\Rightarrow {b^2} = {a^2}\left( {1 - {{\left( {\dfrac{1}{{\sqrt 2 }}} \right)}^2}} \right) = {a^2}\left( {1 - \dfrac{1}{2}} \right) = \dfrac{{{a^2}}}{2}
Now substitute this value in equation (2) we have,
a2a22=1\Rightarrow {a^2} - \dfrac{{{a^2}}}{2} = 1
a22=1\Rightarrow \dfrac{{{a^2}}}{2} = 1
a2=2\Rightarrow {a^2} = 2
a=2\Rightarrow a = \sqrt 2
Now from equation (2), we have
a2b2=1\Rightarrow {a^2} - {b^2} = 1
2b2=1\Rightarrow 2 - {b^2} = 1
b2=1\Rightarrow {b^2} = 1
So the equation of ellipse becomes
x22+y21=1\Rightarrow \dfrac{{{x^2}}}{2} + \dfrac{{{y^2}}}{1} = 1
x2+2y2=2\Rightarrow {x^2} + 2{y^2} = 2
Mow as we know that the foci of an ellipse is given as (±ae1,0)\left( { \pm a{e_1},0} \right)
Therefore, foci of an ellipse = (±2×12,0)=(±1,0)\left( { \pm \sqrt 2 \times \dfrac{1}{{\sqrt 2 }},0} \right) = \left( { \pm 1,0} \right)
So this is the required answer.
Hence options (a) and (b) are the correct answers.

Note: Whenever we face such types of questions the key concept we have to remember is that always recall the condition of orthogonality which is stated above, and always recall the standard equation of ellipse as well as a hyperbola, and always recall the relation between the eccentricity e, semi-major axis a, and semi-minor axis b in case of hyperbola as well as in case of an ellipse.